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Chapter 6: Problem 13

Suppose that \(\mathbf{A}\) is a nonsingular \(n \times n\) matrix and $$\mathbf{U}=\mathbf{F}(\mathbf{X})=\mathbf{A}\left[\begin{array}{c} x_{1}^{2} \\ x_{2}^{2} \\ \vdots \\ x_{n}^{2} \end{array}\right]$$ (a) Show that \(\mathbf{F}\) is regular on the set $$S=\left\\{\mathbf{X} \mid e_{i} x_{i}>0,1 \leq i \leq n\right\\}$$ where \(e_{i}=\pm 1,1 \leq i \leq n\). (b) Find \(\mathbf{F}_{\mathrm{S}}^{-1}(\mathbf{U})\) (c) Find \(\left(\mathbf{F}_{\mathbf{S}}^{-1}\right)^{\prime}(\mathbf{U})\).

Short Answer

Expert verified
When \(\mathbf{A} = I\), the identity matrix, the inverse function becomes: $$\mathbf{F}_S^{-1}(\mathbf{U}) = [\pm \sqrt{U_1}, \pm \sqrt{U_2}, \dots, \pm \sqrt{U_n}]^T.$$ The Jacobian of the inverse function is: $$\left(\mathbf{F}_S^{-1}\right)^{\prime}(\mathbf{U}) = \begin{bmatrix} \frac{e_1}{2\sqrt{U_1}} & 0 & \dots & 0 \\ 0 & \frac{e_2}{2\sqrt{U_2}} & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & \frac{e_n}{2\sqrt{U_n}} \end{bmatrix}.$$

Step by step solution

01

Definition of regular function

A function is regular if and only if it is differentiable and has a non-singular derivative. In our case, we must show that \(\mathbf{F}(\mathbf{X})\) is regular for the given set \(S\). This means that we need to find the Jacobian matrix of \(\mathbf{F}\) and show that it is non-singular for \(\mathbf{X}\) in the set \(S\).
02

Compute the Jacobian of \(\mathbf{F}\)

The Jacobian matrix of \(\mathbf{F}(\mathbf{X})\) is an \(n \times n\) matrix consisting of first-order partial derivatives, given by $$J(\mathbf{F}) = \frac{\partial (\mathbf{A}\mathbf{X}^2)}{\partial \mathbf{X}} = \mathbf{A} \cdot \frac{\partial \mathbf{X}^2}{\partial \mathbf{X}},$$ where $\mathbf{X}^2 = \left[\begin{array}{c} x_{1}^{2} \\ x_{2}^{2} \\ \vdots \\ x_{n}^{2} \end{array}\right]$. Since partial derivative of second-power term is linear, the Jacobian matrix will be as follows: $$J(\mathbf{F}) = \mathbf{A} \begin{bmatrix} 2 x_{1} & 0 & \dots & 0 \\ 0 & 2 x_{2} & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & 2 x_{n} \end{bmatrix}$$
03

Analyze the non-singularity of the Jacobian

Now we show that Jacobian matrix \(J(\mathbf{F})\) is non-singular for all \(\mathbf{X}\) in set \(S\). We know that \(\mathbf{A}\) is a non-singular matrix and \(e_i x_i > 0\), which means that each \(x_i\) is nonzero for \(1 \leq i \leq n\). Therefore, the diagonal matrix with \(2x_i\) as its entries is also non-singular. Since the product of two non-singular matrices is also non-singular, the Jacobian \(J(\mathbf{F})\) is non-singular, and \(\mathbf{F}\) is regular for all \(\mathbf{X}\) in set \(S\). ##Part (b): Find \(\mathbf{F}_S^{-1}(\mathbf{U})\)##
04

Inverse function

Since we have demonstrated that \(\mathbf{F}\) is regular on \(S\), we can now find its inverse function. We need to solve the equation: $$\mathbf{A} \begin{bmatrix} x_{1}^{2} \\ x_{2}^{2} \\ \vdots \\ x_{n}^{2} \end{bmatrix} = \mathbf{U}.$$
05

Solve for \(\mathbf{X}^2\)

To find the inverse function, we first find \(\mathbf{X}^2\). Since \(\mathbf{A}\) is a non-singular matrix, we can multiply both sides by its inverse: $$\mathbf{X}^2 = \mathbf{A}^{-1} \mathbf{U}$$
06

Solve for \(\mathbf{X}\)

Now we can find \(\mathbf{X}\) by taking the element-wise square root of \(\mathbf{X}^2\) and considering the sign of each element \(e_ix_i\). Thus, the inverse function is $$\mathbf{F}_S^{-1}(\mathbf{U}) = [\pm \sqrt{(\mathbf{A}^{-1}\mathbf{U})_1}, \pm \sqrt{(\mathbf{A}^{-1}\mathbf{U})_2}, \dots, \pm \sqrt{(\mathbf{A}^{-1}\mathbf{U})_n}]^T$$ ##Part (c): Find \(\left(\mathbf{F}_S^{-1}\right)^{\prime}(\mathbf{U})\)##
07

Derivative of the inverse function

We have the inverse function \(\mathbf{F}_S^{-1}(\mathbf{U})\). Now we will find its derivative with respect to \(\mathbf{U}\). Since the inverse function is component-wise and element-wise square root applied over \(\mathbf{A}^{-1}\mathbf{U}\), we can compute the Jacobian of \(\mathbf{F}_S^{-1}\) element by element.
08

Compute the derivative of each component

We compute the derivative of the \(i\)-th component of \(\mathbf{F}_S^{-1}(\mathbf{U})\) with respect to \(u_j\): $$\frac{\partial}{\partial u_j} \left(e_i\sqrt{(\mathbf{A}^{-1}\mathbf{U})_i}\right) = \frac{e_i}{2\sqrt{(\mathbf{A}^{-1}\mathbf{U})_i}} \frac{\partial}{\partial u_j} (\mathbf{A}^{-1}\mathbf{U})_i.$$
09

Compute the Jacobian

Notice that $$\frac{\partial}{\partial u_j} (\mathbf{A}^{-1}\mathbf{U})_i = (\mathbf{A}^{-1})_{ij},$$ since matrix multiplication is linear. Therefore, the Jacobian of the inverse function is $$\left(\mathbf{F}_S^{-1}\right)^{\prime}(\mathbf{U}) = \begin{bmatrix} \frac{e_1 (\mathbf{A}^{-1})_{11}}{2\sqrt{(\mathbf{A}^{-1}\mathbf{U})_1}} & \frac{e_1 (\mathbf{A}^{-1})_{12}}{2\sqrt{(\mathbf{A}^{-1}\mathbf{U})_1}} & \dots & \frac{e_1 (\mathbf{A}^{-1})_{1n}}{2\sqrt{(\mathbf{A}^{-1}\mathbf{U})_1}} \\ \frac{e_2 (\mathbf{A}^{-1})_{21}}{2\sqrt{(\mathbf{A}^{-1}\mathbf{U})_2}} & \frac{e_2 (\mathbf{A}^{-1})_{22}}{2\sqrt{(\mathbf{A}^{-1}\mathbf{U})_2}} & \dots & \frac{e_2 (\mathbf{A}^{-1})_{2n}}{2\sqrt{(\mathbf{A}^{-1}\mathbf{U})_2}} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{e_n (\mathbf{A}^{-1})_{n1}}{2\sqrt{(\mathbf{A}^{-1}\mathbf{U})_n}} & \frac{e_n (\mathbf{A}^{-1})_{n2}}{2\sqrt{(\mathbf{A}^{-1}\mathbf{U})_n}} & \dots & \frac{e_n (\mathbf{A}^{-1})_{nn}}{2\sqrt{(\mathbf{A}^{-1}\mathbf{U})_n}} \end{bmatrix}.$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Regular Function
A regular function broadly means that a function is well-behaved, particularly in the sense of being differentiable and having a non-zero, or nonsingular, derivative. In mathematics, ensuring a function is regular is essentially like confirming it continuously changes without any abrupt jumps or holes.

To determine whether a function like \( \mathbf{F}(\mathbf{X}) \) is regular for a given set \( S \), we look at the Jacobian matrix of the function, which encodes information about the function's rate of change.

  • **Differentiability**: A function is differentiable at a point if it has a derivative at that point, meaning it smoothly varies around that point.
  • **Non-singular Derivative**: The derivative (or the matrix in multidimensional cases) of the function should not be singular; a singular matrix is one that does not have an inverse or, equivalently, has a determinant of zero.
Thus, finding the Jacobian and proving it is nonsingular over the set ensures the function \( \mathbf{F} \) remains regular.
Jacobian Matrix
The Jacobian matrix is a critical concept when dealing with functions of multiple variables, indicating how these functions change with respect to their inputs.

In simple terms, for a function \( \mathbf{F}:{\mathbb{R}}^n \to {\mathbb{R}}^m \), the Jacobian matrix is an \( m \times n \) matrix whose \( (i, j) \)th entry is the partial derivative of the \( i \)th output function with respect to the \( j \)th input variable.

  • **Matrix Representation**: For a function \( \mathbf{F}(\mathbf{X}) \), the Jacobian is calculated by differentiating each component of \( \mathbf{X} \). For example, the Jacobian for a function applied to \( \mathbf{A} \mathbf{X}^2 \) involves taking derivatives of quadratic terms, leading often to expressions involving coefficients like \( 2x_i \).
  • **Importance**: The Jacobian matrix is fundamental in determining the behavior of \( \mathbf{F} \). It's crucial for understanding how small changes in \( \mathbf{X} \) affect \( \mathbf{F}(\mathbf{X}) \), as well as ensuring the invertibility of the function; if the Jacobian is non-singular, the function is locally invertible.
This makes the Jacobian matrix a powerful tool for analyzing nonlinear transformations and mapping between different spaces.
Inverse Function
An inverse function essentially reverses the operation of a given function, translating outputs back to their original inputs. For a function \( \mathbf{F} \), the inverse \( \mathbf{F}^{-1} \) satisfies the condition \( \mathbf{F}(\mathbf{F}^{-1}(\mathbf{U})) = \mathbf{U} \) and vice versa.

Determining the inverse of a function involves solving equations that express outputs in terms of inputs, essentially 'reversing' the transformation applied by the function.

  • **Existence**: Not all functions have inverses. For an inverse to exist, a function must be bijective鈥攂oth injective (one-to-one) and surjective (onto).
  • **Finding the Inverse**: In practice, to find the inverse of \( \mathbf{F} \), one needs to express each input, often represented by squared terms like \( x_i^2 \), in terms of the output, and then solve for each \( x_i \).
In the context of problems dealing with matrices, using the inverse of a matrix allows retrieval of the original input from the transformed output, ensuring we can backtrack through the operations performed by \( \mathbf{F} \). Hence, finding the inverse function and proving its differentiability is vital for deeper analysis of the behavior of \( \mathbf{F} \).

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Most popular questions from this chapter

Find \(\mathbf{F}^{\prime}\). (a) \(\mathbf{F}(x, y, z)=\left[\begin{array}{c}(x+y+z) e^{x} \\\ \left(x^{2}+y^{2}\right) e^{-x}\end{array}\right]\) (b) \(\mathbf{F}(x)=\left[\begin{array}{c}g_{1}(x) \\ g_{2}(x) \\ \vdots \\\ g_{n}(x)\end{array}\right]\) (c) \(\mathbf{F}(x, y, z)=\left[\begin{array}{c}e^{x} \sin y z \\ e^{y} \sin x z \\ e^{z} \sin x y\end{array}\right]\)

We say that \(\mathbf{F}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}\) is uniformly continuous on \(S\) if each of its components is uniformly continuous on \(S\). Prove: If \(\mathbf{F}\) is uniformly continuous on \(S,\) then for each \(\epsilon>0\) there is a \(\delta>0\) such that $$ |\mathbf{F}(\mathbf{X})-\mathbf{F}(\mathbf{Y})|<\epsilon \quad \text { if } \quad|\mathbf{X}-\mathbf{Y}|<\delta \quad \text { and } \quad \mathbf{X}, \mathbf{Y} \in S. $$

Let \(\mathbf{A}\) be an \(m \times n\) matrix. (a) Use Exercises 6.2 .7 and 6.2 .8 to show that the quantitites $$ M(\mathbf{A})=\max \left\\{\frac{|\mathbf{A} \mathbf{X}|}{|\mathbf{X}|} \mid \mathbf{X} \neq \mathbf{0}\right\\} \quad \text { and } \quad m(\mathbf{A})=\min \left\\{\frac{|\mathbf{A} \mathbf{X}|}{|\mathbf{X}|} \mid \mathbf{X} \neq \mathbf{0}\right\\} $$ exist. HINT: Consider the function \(\mathbf{L}(\mathbf{Y})=\mathbf{A} \mathbf{Y}\) on \(S=\\{\mathbf{Y}|| \mathbf{Y} \mid=1\\}\). (b) Show that \(M(\mathbf{A})=\|\mathbf{A}\|\). (c) Prove: If \(n>m\) or \(n=m\) and \(\mathbf{A}\) is singular, then \(m(\mathbf{A})=0\). (This requires a result from linear algebra on the existence of nontrivial solutions of \(\mathbf{A X}=\mathbf{0}\).) (d) Prove: If \(n=m\) and \(\mathbf{A}\) is nonsingular, then $$ m(\mathbf{A}) M\left(\mathbf{A}^{-1}\right)=m\left(\mathbf{A}^{-1}\right) M(\mathbf{A})=1 $$

Let \(u, v,\) and \(x\) be continuously differentiable functions of \((w, y)\) that satisfy $$ \begin{aligned} x^{2} y+x y^{2}+u^{2}-(v+w)^{2} &=-3 \\ e^{x+y}-u-v-w &=-2 \\ (x+y)^{2}+u+v+w^{2} &=3 \end{aligned} $$ near \(\left(w_{0}, y_{0}\right)=(0,-1),\) and suppose that $$ u(0,-1)=1, \quad v(0,-1)=2, \quad x(0,-1)=1 $$ Find the first partial derivatives of \(u, v,\) and \(x\) with respect to \(y\) and \(w\) at (0,-1) .

Show that if \(\mathbf{F}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}\) is differentiable at \(\mathbf{X}_{0}\) and \(\epsilon>0,\) there is a \(\delta>0\) such that $$ \left|\mathbf{F}(\mathbf{X})-\mathbf{F}\left(\mathbf{X}_{0}\right)\right| \leq\left(\left\|\mathbf{F}^{\prime}\left(\mathbf{X}_{0}\right)\right\|+\epsilon\right)\left|\mathbf{X}-\mathbf{X}_{0}\right| \quad \text { if } \quad\left|\mathbf{X}-\mathbf{X}_{0}\right|<\delta . $$ Compare this with Lemma \(6.2 .5 .\).
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